arXiv:math/0601425v1 [math.QA] 18 Jan 2006
UNITARY REPRESENTATIONS OF THE EXTENDED AFFINE LIE ALGEBRA gl^ (Cq ) 3
Ziting Zeng Abstract. We present modules for the extended affine Lie algebra gl^ 3 (Cq ) by using the idea of free fields. A necessary and sufficient condition for the modules being unitary is given.
§0. Introduction. Let q be a non-zero complex number. A quantum 2-torus associated to q (see [M]) is the unital associative C-algebra Cq [s±1 , t±1 ] (or, simply Cq ) with generators s±1 , t±1 and relations ss−1 = s−1 s = tt−1 = t−1 t = 1 and ts = qst.
(0.1)
Define κ, : Cq → C to be a C-linear function given by κ(sm tn ) = δ(m,n),(0,0)
(0.2)
Let ds , dt be the degree operators on Cq defined by ds (sm tn ) = msm tn , dt (sm tn ) = nsm tn
(0.3) for m, n ∈ Z.
Let gl3 (Cq ) be the Lie algebra of 3 by 3 matrices whose entries are from Cq . We form a natural central extension of gl3 (Cq ) as follows. (0.4)
gl\ 3 (Cq ) = gl3 (Cq ) ⊕ Ccs ⊕ Cct
with Lie bracket (0.5) [Eij (sm1 tn1 ), Ekl (sm2 tn2 )] =δjk q n1 m2 Eil (sm1 +m2 tn1 +n2 ) − δil q n2 m1 Ekj (sm1 +m2 tn1 +n2 ) + m1 q n1 m2 δjk δil δm1 +m2 ,0 δn1 +n2 ,0 cs + n1 q n1 m2 δjk δil δm1 +m2 ,0 δn1 +n2 ,0 ct Typeset by AMS-TEX
1
for m1 , m2 , n1 , n2 ∈ Z, 1 ≤ i, j, k, l ≤ 3, where Eij is the matrix whose (i, j)-entry is 1 and 0 elsewhere, and cs and ct are central elements of gl\ 3 (Cq ). The derivations ds and dt can be extended to derivations on gl3 (Cq ). Now we can define the semi-direct product of the Lie algebra gl\ 3 (Cq ) and those derivations: (0.6)
\ gl^ 3 (Cq ) = gl3 (Cq ) ⊕ Cds ⊕ Cdt .
The Lie algebra gl^ 3 (Cq ) is an extended affine Lie algebra of type A2 with nullity 2. (See [AABGP] and [BGK] for definitions). Extended affine Lie algebras are a higher dimensional generalization of affine KacMoody Lie algebras introduced by [H-KT] and systematically studied in [AABGP] and [BGK]. It turns out that any extended affine Lie algebra of type A is coordinated by a quantum torus (or a nonassociative torus for some small rank cases). Representations for extended affine Lie algebras coordinated by quantum tori and Lie algebras related to quantum tori have been studied in [JK2], [BS], [G1,2,3], [ER1,2], [EB],[GZ] [EZ], [LT1,2], [G-KK], [VV], [Mi], [ZZ], [BZ], [SZ], [L] and [BEG], and among others. The Wakimoto’s free fields construction provides a remarkable way to realize affine Kac-Moody Lie algebras (see [W2], [FF] and [EFK]). In [GZ], we used Wakimoto’s idea to construct a class of representations for gl^ 2 (Cq ) and found out the necessary and sufficient condition for the representations being unitary. In this paper, we will continue to ^ construct representations for gl^ 3 (Cq ). As witnessed in [FF], the realization for gl3 (Cq ) is much more subtle and complicated than the one for gl^ 2 (Cq ). We then go on to construct a hermitian form and to determine when the form is positive definite (so the representations are unitary). Unlike [GZ] in which we defined the form on the monomial basis for the module (this idea goes back to [W1]), we define the form directly on the basis consisting of certain iterated module actions on a “highest weight vector” 1. This facilitates the verification of the defined form being a hermitian from. Throughout this paper, we denote the field of complex numbers, real numbers and the ring of integers by C, R and Z respectively. §1. Module for gl^ 3 (Cq ).
In this section, we use Wakimoto’s idea [W1] to construct gl^ 3 (Cq )-modules as was done in [GZ]. Let K1 = {(3m + 1, 3n + 1), m, n ∈ Z}, and K−1 = {(3m − 1, 3n − 1), m, n ∈ Z}. If A = (3m + 1, 3n + 1) ∈ K1 , we always write A1 = m, A2 = n, and similarly, if B = (3m − 1, 3n − 1) ∈ K−1 , then B1 = m, B2 = n. Set (1.1)
V = C[xA , xB : A ∈ K1 , B ∈ K−1 ] 2
be the (commutative) polynomial ring of infinitely many variables. The operators x(m,n) ∂ and ∂x(m,n) act on V as the usual multiplication and differentiation operators respectively. Given a family of 2 × 2 lower triangular matrices a(m,n) 0 ∈ SL2 (C) Xm,n = c(m,n) d(m,n) S for (m, n) ∈ K1 K−1 (so a(m,n) d(m,n) = 1). Set ∂ P(m,n) = a(m,n) (1.2) ∂x(m,n) ∂ Q(m,n) = c(m,n) (1.3) + d(m,n) x(m,n) ∂x(m,n) S for (m, n) ∈ K1 K−1 . Then for A, A′ ∈ K1 , B, B ′ ∈ K−1 ,
[PA , PA′ ] = [QA , QA′ ] = [PB , PB ′ ] = [QB , QB ′ ] = 0 [PA , PB ] = [PA , QB ] = [QA , QB ] = [PB , QA ] = 0 [PA , QA′ ] = δA,A′ , [PB , QB ′ ] = δB,B ′ .
Fix a complex number µ, we define the following operators on V : (1.4) (µ)
e21 (m1 , n1 ) = − q −m1 n1 µP−(3m1 −1,3n1 −1) X ′ ′ − q n1 A1 +A2 m1 +A2 A1 QA+A′ +(3m1 −1,3n1 −1) PA PA′ A,A′ ∈K1
−
X
q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB
A∈K1 B∈K−1
(1.5) (µ)
e12 (m1 , n1 ) =Q(3m1 +1,3n1 +1) (1.6) (µ)
e11 (m1 , n1 ) =
X
A∈K1
(1.7) (µ)
e22 (m1 , n1 ) = −
1 q A1 n1 Q(3m1 ,3n1 )+A PA + µδ(m1 ,n1 ),(0,0) 2
X
q A2 m1 Q(3m1 ,3n1 )+A PA
A∈K1
−
X
B∈K−1
1 q B2 m1 Q(3m1 ,3n1 )+B PB − µδ(m1 ,n1 ),(0,0) 2 3
(1.8) (µ)
e23 (m1 , n1 ) = −q −m1 n1 µP−(3m1 +1,3n1 +1) X q n1 B1 +A2 m1 +A2 B1 QA+B+(3m1 +1,3n1 +1) PA PB − A∈K1 B∈K−1
−
X
′
′
q n1 B1 +B2 m1 +B2 B1 QB+B ′ +(3m1 +1,3n1 +1) PB PB ′
B,B ′ ∈K−1
(1.9) (µ)
e32 (m1 , n1 ) = Q(3m1 −1,3n1 −1) (1.10) (µ)
e31 (m1 , n1 ) =
X
q A1 n1 Q(3m1 −2,3n1 −2)+A PA
A∈K1
(1.11) (µ)
e13 (m1 , n1 ) =
X
q B1 n1 Q(3m1 +2,3n1 +2)+B PB
B∈K−1
(1.12) (µ)
(µ)
(1.13)
D1
(1.14)
(µ) D2
X
1 q B1 n1 Q(3m1 ,3n1 )+B PB + µδ(m1 ,n1 ),(0,0) 2 B∈K−1 X X = A 1 Q A PA + B1 QB PB
e33 (m1 , n1 ) =
A∈K1
=
X
B∈K−1
A 2 Q A PA +
A∈K1
X
B2 QB PB
B∈K−1
Although the operators are infinite sums, they are well-defined as operators on V . Now we have the following result: Theorem 1.15. The linear map π : gl^ 3 (Cq ) → End V given by (µ)
π(Eij (sm1 tn1 )) = eij (m1 , n1 ), (µ)
π(ds ) = D1 ,
(µ)
π(dt ) = D2 ,
π(cs ) = π(ct ) = 0,
for m1 , n1 ∈ Z, 1 ≤ i, j ≤ 3, is a Lie algebra homomorphism. Proof. The proof is straightforward. However, we would like to provide a few details. We shall do this orderly so that we won’t miss any case. 4
First, we have (µ)
(µ)
[e11 (m1 , n1 ), e11 (m2 , n2 )] X X ′ =[ q A1 n1 Q(3m1 ,3n1 )+A PA , q A1 n1 Q(3m2 ,3n2 )+A′ PA′ ] A′ ∈K1
A∈K1
=
X
′ ′ 1 q (m2 +A1 )n1 +A n2 Q(3m1 ,3n1 )+(3m2 +3n2 )+A′ PA′ + q m2 n1 µδ(m1 +m2 ,n1 +n2 ),(0,0) 2 A′ ∈K1 X 1 − q A1 n1 +(m1 +A1 )n2 q(3m1 ,3n1 )+(3m2 ,3n2 )+A PA − q m1 n2 µδ(m1 +m2 ,n1 +n2 ),(0,0) 2
A∈K1
(µ)
(µ)
=q m2 n1 e11 (m1 + m2 , n1 + n2 ) − q m1 n2 e11 (m1 + m2 , n1 + n2 ). The following two brackets are easy. (µ)
(µ)
(µ)
(µ)
(µ)
(µ)
[e11 (m1 , n1 ), e12 (m2 , n2 )] = q m2 n1 e12 (m1 + m2 , n1 + n2 ) [e11 (m1 , n1 ), e13 (m2 , n2 )] = q m2 n1 e13 (m1 + m2 , n1 + n2 ). (µ)
(µ)
[e11 (m1 , n1 ), e21 (m2 , n2 )] X = (−µ)q A1 n1 −m2 n2 [Q(3m1 ,3n1 )+A PA , P−(3m2 −1,3n2 −1) ] A∈K1
−
X
¯′
¯
¯ ¯′
q A1 n1 +A1 n2 +A2 m2 +A2 A1 [Q(3m1 ,3n1 )+A PA , QA+ ¯ ′ +(3m2 −1,3n2 −1) PA ¯′ ] ¯ PA ¯ A
A∈K1 ¯A ¯ ′ ∈K1 A,
−
X
¯
¯
q A1 n1 +n2 A1 +B2 m2 +B2 A1 [Q(3m1 ,3n1 )+A PA , QA+B+(3m P ¯P ] ¯ 2 −1,3n2 −1) A B
A∈K1 ¯ A∈K 1 B∈K−1
=µq −(m1 +m2 )(n1 +n2 )+m1 n2 P−(3(m1 +m2 )−1,3(n1 +n2 )−1) X ¯ ¯′ ¯′ ¯ ¯ ¯′ q (A1 +A1 +m2 )n1 +n2 A1 +A2 m2 +A2 A1 Q(3m1 ,3n1 )+A+ − ¯ A ¯′ +(3m2 −1,3n2 −1) PA ¯ PA ¯′ ¯A ¯ ′ ∈K1 A,
+
X
¯
¯
P ¯P q A1 n1 +n2 (m1 +A1 )+A2 m2 +A2 (m1 +A1 ) QA+(3m ¯ 1 ,3n1 )+A+(3m2 −1,3n2 −1) A A
¯ A,A∈K 1
+
X
¯′
′
q A1 n1 +n2 A1 +(n1 +A2 )m2 +(n1 +A2 )A1 QA+(3m1 ,3n1 )+A¯′ +(3m2 −1,3n2 −1) PA¯′ PA
¯ ′ ∈K1 A,A
+
X
q A1 n1 +n2 (m1 +A1 )+bdb2 m2 +B2 (m1 +A1 ) QA+(3m1 ,3n1 )+B+(3m2 −1,3n2 −1) PB PA
A∈K1 B∈K−1
5
(the second term and the fourth term are negative to each other) = − q m1 n2 (−µq −(m1 +m2 )(n1 +n2 ) P−(3(m1 +m2 )−1,3(n1 +n2 )−1) X ¯ ¯ − q A(n1 +n2 )+A2 (m1 +m2 )+A1 A2 Q(3(m1 +m2 )−1,3(n1 +n2 )−1)+A+A¯ PA¯ PA ¯ A,A∈K 1
−
X
q A1 (n1 +n2 )+B2 (m1 +m2 )+B2 +A1 QA+B+(3(m1 +m2 )−1,3(n1 +n2 )−1) PB PA )
A∈K1 B∈K−1 (µ)
= − q m1 n2 e21 (m1 + m2 , n1 + n2 ). The following seven brackets are easily verified. (µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
[e11 (m1 , n1 ), e22 (m2 , n2 )] = 0 [e11 (m1 , n1 ), e23 (m2 , n2 )] = 0 (µ)
[e11 (m1 , n1 ), e31 (m2 , n2 )] = −q m1 n2 e31 (m1 + m2 , n1 + n2 ) [e11 (m1 , n1 ), e32 (m2 , n2 )] = 0 [e11 (m1 , n1 ), e33 (m2 , n2 )] = 0 [e12 (m1 , n1 ), e12 (m2 , n2 )] = 0 [e12 (m1 , n1 ), e13 (m2 , n2 )] = 0. Next, we have (µ)
(µ)
[e12 (m1 , n1 ), e21 (m2 , n2 )] =µq −m2 n2 δ(m1 ,n1 ),(−m2 ,n2 ) X ′ ′ + q n2 A1 +n1 m2 +n1 A1 Q(3m1 +1,3n1 +1)+A′ +(3m2 −1,3n2 −1)PA′ A′ ∈K1
+
X
q n2 m1 +A2 m2 +A2 m1 QA+(3m1 +1,3n1 +1)+(3m2 −1,3n2 −1) PA
A∈K1
+
X
q n2 m1 +B2 m2 +B2 m1 Q(3m1 +1,3n1 +1)+B+(3m2 −1,3n2 −1) PB
B∈K−1
6
X
′ 1 q (n1 +n2 )A1 Q(3m1 +3m2 ,3n1 +3n2 )+A′ PA′ + δ(m1 ,n1 ),(−m2 ,n2 ) ) 2 A′ ∈K1 X − q n2 m1 (− q A2 (m2 +m1 ) QA+(3m1 +3m2 ,3n1 +3n2 ) PA
=q n1 m2 (
−
A∈K1
X
1 q B2 (m2 +m1 ) Q(3m1 +3m2 ,3n1 +3n2 )+B PB − δ(m1 ,n1 ),(−m2 ,n2 ) ) 2
B∈K−1 (µ)
(µ)
=q n1 m2 e11 (m1 + m2 , n1 + n2 ) − q n2 m1 e22 (m1 + m2 , n1 + n2 ).
The following six brackets can be checked easily.
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
[e12 (m1 , n1 ), e22 (m2 , n2 )] = q n1 m2 e12 (m1 + m2 , n1 + n2 ) [e12 (m1 , n1 ), e23 (m2 , n2 )] = q n1 m2 e13 (m1 + m2 , n1 + n2 ) (µ)
[e12 (m1 , n1 ), e31 (m2 , n2 )] = −q m1 n2 e32 (m1 + m2 , n1 + n2 ) [e12 (m1 , n1 ), e32 (m2 , n2 )] = 0 [e12 (m1 , n1 ), e33 (m2 , n2 )] = 0 [e13 (m1 , n1 ), e13 (m2 , n2 )] = 0.
[e13 (m1 , n1 ), e21 (m2 , n2 )] X q B1 n1 Q(3m1 +2,3n1 +2)+B PB , −q −m2 n2 µP−(3m2 −1,3n2 −1) =[ B∈K−1
−
X
′
′
q n2 A1 +A2 m2 +A2 A1 QA+A′ +(3m2 −1,3n2 −1) PA PA′
A,bda′ ∈K1
−
X
q n2 A1 +B2 m2 +B2 A1 QA+B+(3m2 −1,3n2 −1) PA PB ]
A∈K1 B∈K−1
7
=−
X
q (A1 +B1 +m2 )n1 +n2 +A1 +B2 m2 +B2 A1 Q(3m1 +2,3n1 +2)+A+B+(3m2 −1,3n2 −1) PA PB
A∈K1 B∈K−1
+ q −m2 n2 +(−m1 −m2 )n1 µP(−3m1 −3m2 −1,−3n1 −3n2 −1) X + q n2 (m1 +B1 )+A2 m2 +A2 (m1 +B1 )+B1 n1 QA+(3m1 +2,3n1 +2)+B+(3m2 −1,3n2 −1) PA PB A∈K1 B∈K−1
+
X
′
′
q n2 A1 +(n1 +B2 )m2 +(n1 +B2 )A1 +B1 n1 QA′ +(3m1 +2,3n1 +2)+B+(3m2 −1,3n2 −1) PA′ PB
A′ ∈K1 B∈K−1
+
X
′
′
′
q n2 (B1 +m1 )+B2 m2 +B2 (B1 +m1 )+B1 n1 QB ′ +(3m1 +2,3n1 +2)+B+(3m2 −1,3n2 −1) PB ′ PB
B,B ′ ∈K−1
(the first and the fourth terms cancel each other) =q n2 m1 (q −(m1 +m2 )(n1 +n2 ) µP−(3m1 +3m2 −1,3n1 +3n2 −1) X ′ ′ + q (n1 +n2 )A1 +A2 (m1 +m2 )+A2 A1 QA+A′ +(3m1 +3m2 −1,3n1 +3n2 −1) PA PA′ A,bda′ ∈K1
−
X
q (n1 +n2 )A1 +B2 (m1 +m2 )+B2 A1 QA+B+(3m1 +3m2 −1,3n1 +3n2 −1) PA PB )
A∈K1 B∈K−1 (µ)
= − q n2 m1 e23 (m1 + m2 , n1 + n2 ). The following two brackets are easy. (µ)
(µ)
(µ)
(µ)
[e13 (m1 , n1 ), e22 (m2 , n2 )] = 0 [e13 (m1 , n1 ), e23 (m2 , n2 )] = 0.
(µ)
(µ)
(µ)
(µ)
− [e13 (m2 , n2 ), e31 (m1 , n1 )] = [e31 (m1 , n1 ), e13 (m2 , n2 )] X q n1 A1 +n2 B1 [Q(3m1 −2,3n1 −2)+A PA , Q(3m2 +2,3n2 +2)+B PB ] = A∈K1 B∈K−1
=
X
q n2 B1 +n1 (m2 +B1 ) Q(3m1 +3m2 ,3n1 +3n2 )+B PB
B∈K−1
−
X
q n1 A1 +n2 (m1 +A1 ) Q(3m1 +3m2 ,3n1 +3n2 )+A PA
A∈K1
8
X
=q n1 m2 (
B∈K−1
− q n2 m1 (
1 q n2 B1 +n1 (m2 +B1 ) Q(3m1 +3m2 ,3n1 +3n2 )+B PB + µδ(m1 +m2 ,n1 +n2 ),(0,0) ) 2
X
A∈K1 (µ) =q n1 m2 e33 (m1
1 q n1 A1 +n2 (m1 +A1 ) Q(3m1 +3m2 ,3n1 +3n2 )+A PA + µδ(m1 +m2 ,n1 +n2 ),(0,0) ) 2 (µ)
+ m2 , n1 + n2 ) − q n2 m1 e11 (m1 + m2 , n1 + n2 ).
The following two brackets are easy. (µ)
(µ)
(µ)
(µ)
(µ)
(µ)
[e13 (m1 , n1 ), e32 (m2 , n2 )] = q n1 m2 e12 (m1 + m2 , n1 + n2 ) [e13 (m1 , n1 ), e33 (m2 , n2 )] = q n1 m2 e13 (m1 + m2 , n1 + n2 ). (µ)
(µ)
[e21 (m1 , n1 ), e21 (m2 , n2 )] X =µq −m1 n1 q (−m1 −m2 −A1 )n2 +A2 m2 +A2 (−m1 −m2 −A1 ) PA P(−3m1 −3m2 ,−3n1 −3n2 )−A A∈K1
− µq −m2 n2
′
′
′
′
q n1 A1 +(−n1 −n2 −A2 )m1 +(−n1 −n2 −A2 )A1 PA′ P(−3m1 −3m2 ,−3n1 −3n2 )−A′
A′ ∈K1
X
+[
X
′
′
q n1 A1 +A2 m1 +A2 A1 QA+A′ +(3m1 −1,3n1 −1) PA PA′ ,
A,A′ ∈K1
X
′
′
q n2 A1 +A2 m2 +A2 A1 QA+A′ +(3m2 −1,3n2 −1) PA PA′ ]
A,A′ ∈K1
+[
X
′
′
q n1 A1 +A2 m1 +A2 A1 QA+A′ +(3m1 −1,3n1 −1) PA PA′ ,
A,A′ ∈K1
X
q n2 A1 +B2 m2 +B2 A1 QA+B+(3m2 −1,3n2 −1) PA PB ]
A∈K1 B∈K−1
+[
X
q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB ,
A∈K1 B∈K−1
X
′
′
q n2 A1 +A2 m2 +A2 A1 QA+A′ +(3m2 −1,3n2 −1) PA PA′ ]
A,A′ ∈K1
+[
X
q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB ,
A∈K1 B∈K−1
X
q n2 A1 +B2 m2 +B2 A1 QA+B+(3m2 −1,3n2 −1) PA PB ]
A∈K1 B∈K−1
9
(the first and the second terms cancel each other) X
=
¯
¯′
¯
¯′
¯′
¯
¯ ¯′
q n1 (A1 +A1 +m2 )+A2 m1 +A2 (A1 +A1 +m2 )+n2 A1 +A2 m2 +A2 A1
¯A ¯ ′ ∈K1 A,A,
.QA+A+ ¯ A ¯′ +(3m1 −1,3n1 −1)+(3m2 −1,3n2 −1) PA PA ¯ PA ¯′ X ′ ′ ′ ′ ¯ ¯ ¯ ¯ ¯′ ¯ ¯ ¯′ + q n1 A1 +(A2 +A2 +n2 )m1 +(A2 +A2 +n2 )A1 +n2 A1 +A2 m2 +A2 A1 ¯A ¯ ′ ∈K1 A′ ,A,
.QA′ +A+ ¯′ ¯ PA ¯ A ¯′ +(3m1 −1,3n1 −1)+(3m2 −1,3n2 −1) PA′ PA
−
X
′
¯
¯
′
′
′
q n2 (A1 +A1 +m1 )+A2 m2 +A2 (A1 +A1 +m1 )+n1 A1 +A2 m1 +A2 A1
¯ A,A′ ,A∈K 1
.QA+A′ +A+(3m P P ′ PA¯ ¯ 1 −1,3n1 −1)+(3m2 −1,3n2 −1) A A X ′ ′ ′ ′ ′ ¯ ¯′ q n2 A1 +(A2 +A2 +n1 )m2 +(A2 +A2 +n1 )A1 +n1 A1 +A2 m1 +A2 A1 − ¯ ′ ∈K1 A,A′ ,A
.QA+A′ +A¯′ +(3m1 −1,3n1 −1)+(3m2 −1,3n2 −1) PA PA′ PA¯′ X ′ ′ ′ ′ − q n2 (A1 +A1 +m1 )+B2 m2 +B2 (A1 +A1 +m1 )+n1 A1 +A2 m1 A2 A1 A,A′ ∈K1 B∈K−1
.QA+A′ +B+(3m1 −1,3n1 −1)+(3m2 −1,3n2 −1) PA PA′ PB X ′ ′ ′ ′ + q n1 (A1 +A1 +m2 )+B2 m1 +B2 (A1 +A1 +m2 )+n2 A1 +A2 m2 +A2 A1 A,A′ ∈K1 B∈K−1
.QA+A′ +B+(3m1 −1,3n1 −1)+(3m2 −1,3n2 −1) PA PA′ PB X ¯ ¯ ¯ ¯ + q n1 A1 +(A2 +B2 +n2 )m1 +(A2 +B2 +n2 )A1 +n2 A1 +B2 m2 +B2 A1 ¯ A,A∈K 1 B∈K−1
.QA+A+B+(3m P P ¯P ¯ 1 −1,3n1 −1)+(3m2 −1,3n2 −1) A A B X ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ − q n2 A1 +(A2 +B2 +n1 )m2 +(A2 +B2 +n1 )A1 +n1 A1 +B2 m1 +B2 A1 ¯ A,A∈K 1 ¯ B∈K −1
.QA+A+ P P ¯P ¯ = 0 ¯ B+(3m ¯ 1 −1,3n1 −1)+(3m2 −1,3n2 −1) A A B as the first term cancels the fourth, the second term cancels the third, the fifth term cancels the seventh, and the sixth term cancels the eighth. 10
(µ)
(µ)
(µ)
(µ)
− [e21 (m2 , n2 ), e22 (m1 , n1 )] = [e22 (m1 , n1 ), e21 (m2 , n2 )] X =µ q A2 m1 −m2 n2 [Q(3m1 ,3n1 )+A PA , P−(3m2 −1,3n2 −1) ] A∈K1
+
X
¯
′
¯
′
q A2 m1 +n2 A1 +A2 m2 +A2 A1 [Q(3m1 ,3n1 )+A PA , QA+A ′ +(3m −1,3n −1) PA′ PA ¯ ¯] 2 2
¯ A,A′ ,A∈K 1
+
X
¯
¯
P P ] q A2 m1 +n2 A1 +B2 m2 +B2 A1 [Q(3m1 ,3n1 )+A PA , QA+B+(3m ¯ 2 −1,3n2 −1) A B
¯ A,A∈K 1 B∈K−1
+
X
¯
¯
q B2 m1 +n2 A1 +B2 m2 +B2 A1 [Q(3m1 ,3n1 )+B PB , QA+B+(3m P P ¯] ¯ 2 −1,3n2 −1) A B
A∈K1 ¯ B,B∈K −1
= − µq −m2 n2 +(−n1 −n2 )m1 P−3m1 −3m2 +1,−3n1 −3n2 +1 X ′ ′ ′ + q (A2 +A2 +n2 )m1 +n2 A1 +A2 m2 +A2 A1 Q(3m1 ,3n1 )+A+A′ +(3m2 −1,3n2 −1) PA PA′ A,A′ ∈K1
−
X
¯
¯
¯
q n2 (m1 +A1 )+A2 m2 +A2 (m1 +A1 )+A2 m1 QA+A+(3m P P¯ ¯ 1 ,3n1 )+(3m2 −1,3n2 −1) A A
¯ A,A∈K 1
−
X
′
¯
¯
′
¯
q n2 A1 +(A2 +n1 )m2 +(A2 +n1 )A1 +A2 m1 QA′ +A+(3m P ′ PA¯ ¯ 1 ,3n1 )+(3m2 −1,3n2 −1) A
¯ A′ ,A∈K 1
−
X
q n2 (A1 +m1 )+B2 m2 +B2 (A1 +m1 )+A2 m1 QA+B+(3m1 ,3n1 )+(3m2 −1,3n2 −1) PA PB
X
q (A2 +B2 +n2 )m1 +n2 A1 +B2 m2 +B2 A1 QA+B+(3m1 ,3n1 )+(3m2 −1,3n2 −1) PA PB
X
q n2 A1 +(n1 +B2 )m2 +(n1 +B2 )A1 +B2 m1 QA+B+(3m1 ,3n1 )+(3m2 −1,3n2 −1) PA PB
A∈K1 B∈K−1
+
A∈K1 B∈K−1
−
A∈K1 B∈K−1
(the second and the third terms are cancelled, and the fifth and the sixth terms are 11
cancelled)
=q n1 m2 (−µq −(n1 +n2 )(m1 +m2 ) P−(3(m1 +m2 )−1,3(n1 +n2 )−1) X ′ ′ ¯ ¯ q (n1 +n2 )A1 +(m1 +m2 )A2 +A2 A1 Q(3m1 +3m2 −1,3n1 +3n2 −1)+A′ +A¯ PA′ PA¯ − ¯ ′ ∈K1 A,A
X
−
q (n1 +n2 )A1 +(m1 +m2 )B2 +B2 A1 )Q(3m1 +3m2 −1,3n1 +3n2 −1)+A+B PA PB
A∈K1 B∈K−1 (µ)
=q n1 m2 e21 (m1 + m2 , n1 + n2 ).
(µ)
(µ)
[e21 (m1 , n1 ), e23 (m2 , n2 )] X =µq −m1 n1 q n2 (−m1 −m2 −A1 )+A2 m2 +A2 (−m1 −m2 −A1 ) PA P(−3m1 −3m2 ,−3n1 −3n2 )−A +[
X
A∈K1
′
′
q n1 A1 +A2 m1 +A2 A1 QA+A′ +(3m1 −1,3n1 −1) PA PA′ ,
A,A′ ∈K1
X
q n2 B1 +A2 m2 +A2 B1 Q(3m2 +1,3n2 +1)+A+B PA PB ]
A∈K1 B∈K−1
− µq −m2 n2 +[
X
X
q n1 A1 +(−n1 −n2 −A2 )m1 +(−n1 −n2 −A2 )A1 PA P(−3m1 −3m2 ,−3n1 −3n2 )−A
A∈K1
q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB ,
A∈K1 B∈K−1
X
q n2 B1 +A2 m2 +A2 B1 Q(3m2 +1,3n2 +1) PA PB ]
A∈K1 B∈K−1
+[
X
q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB ,
A∈K1 B∈K−1
X
′
′
q n2 B1 +B2 m2 +B2 B1 Q(3m2 +1,3n2 +1)+B+B ′ PB PB ′ ]
B,B ′ ∈K−1
12
(the first term and the third term are negative to each other) X ¯ ¯ ¯ ¯ = q n1 (m2 +A1 +B1 )+A2 m1 +A2 (m1 +A1 +B1 )+n2 B1 +A2 m2 +A2 B1 ¯ A,A∈K 1 B∈K−1
.QA+A+B+(3m P P ¯P ¯ 1 +3m2 ,3n1 +3n2 ) A A B X ′ ¯ ¯ ¯ ¯ ¯ + q n1 A1 +(n2 +A2 +B2 )m1 +(n2 +A2 +B2 )A1 +n2 B1 +A2 m2 +A2 B1 ¯ A′ ,A∈K 1 B∈K−1
.QA′ +A+B+(3m P ′ PA¯ PB ¯ 1 +3m2 ,3n1 +3n2 ) A X ′ ′ ′ ′ − q n2 B1 +(A2 +A2 +n1 )m2 +(A2 +A2 +n1 )B1 +n1 A1 +A2 m1 +A2 A1 A,A′ ∈K1 B∈K−1
.QA′ +A+B+(3m1 +3m2 ,3n1 +3n2 ) PA′ PA PB +
X
′
′
′
′
q n1 (m2 +A1 +B1 )+B2 m1 +B2 (m2 +A1 +B1 )+n2 B1 +A2 m2 +A2 B1
A∈K1 B,B ′ ∈K−1
.QA+B+B ′ +(3m1 +3m2 ,3n1 +3n2 ) PA PB PB ′ X ¯ ¯ − q n2 (A1 +B1 +m1 )+A2 m2 +A2 (A1 +B+m1 )+n1 A1 +B2 m1 +B2 A1 ¯ A,A∈K 1 B∈K−1
.QA+A+B+(3m P P ¯P ¯ 1 +3m2 ,3n1 +3n2 ) A A B X ′ ′ ′ ′ + q n1 A1 +(n2 +B2 +B2 )m1 +(n2 +B2 +B2 )A1 +n2 B1 +B2 m2 +B2 B1 A∈K1 B,B ′ ∈K−1
.QA+B+B ′ +(3m1 +3m2 ,3n1 +3n2 ) PA PB PB ′ X ¯ ¯ ¯ ¯ − q n2 (A1 +B1 +m1 )+B2 m2 +B2 (A1 +B1 +m1 )+n1 A1 +B2 m1 +B2 A1 A∈K1 ¯ B,B∈K −1
.QA+B+B+(3m P P P¯ ¯ 1 +3m2 ,3n1 +3n2 ) A B B X ′ ′ ¯ ¯ ¯ ¯ q n2 B1 +(A2 +B2 +n1 )m2 +(A2 +B2 +n1 )B1 +n1 A1 +B2 m1 +B2 A1 − A∈K1 ¯ B ′ ,B∈K −1
.QA+B ′ +B+(3m P P ′ PB¯ = 0 ¯ 1 +3m2 ,3n1 +3n2 ) A B as the first term and the third, the second and the fifth, the fourth and the eighth, the sixth and the seventh are cancelled. 13
The following three brackets are easy.
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
[e21 (m1 , n1 ), e31 (m2 , n2 )] = 0 (µ)
[e21 (m1 , n1 ), e32 (m2 , n2 )] = −q n2 m1 e31 (m1 + m2 , n1 + n2 ) [e21 (m1 , n1 ), e33 (m2 , n2 )] = 0.
(µ)
(µ)
[e22 (m1 , n1 ), e22 (m2 , n2 )] X ′ = q A2 m1 +A2 m2 [Q(3m1 ,3n1 )+A PA , Q(3m2 ,3n2 )+A′ PA′ ] A,A′ ∈K1
+
X
′
q B2 m1 +B2 m2 [Q(3m1 ,3n1 )+B PB , Q(3m2 ,3n2 )+B ′ PB ′ ]
B,B ′ ∈K−1
=
X
′
′
q (n2 +A2 )m1 +A2 m2 Q(3m1 ,3n1 )+(3m2 ,3n2 )+A′ PA′
A′ ∈K1
−
X
q A2 m1 +(n1 +A2 )m2 Q(3m1 ,3n1 )+(3m2 ,3n2 )+A PA
A∈K1
+
X
′
′
q (n2 +B2 )m1 +B2 m2 Q(3m1 ,3n1 )+(3m2 ,3n2 )+B ′ PB ′
B ′ ∈K−1
−
X
q B2 m1 +(n1 +B2 )m2 Q(3m1 ,3n1 )+(3m2 ,3n2 )+B PB
B∈K−1
=q n1 m2 (− −
X
−
X
X
q A2 (m1 +m2 ) Q(3m1 ,3n1 )+(3m2 ,3n2 )+A PA
A∈K1
1 q B2 (m1 +m2 ) Q(3m1 ,3n1 )+(3m2 ,3n2 )+B PB − µδ(m1 +m2 ,n1 +n2 ),(0,0) ) 2 B∈K−1 X − q n2 m1 (− q A2 (m1 +m2 ) Q(3m1 ,3n1 )+(3m2 ,3n2 )+A PA B∈K−1 (µ)
A∈K1
1 q B2 (m1 +m2 ) Q(3m1 ,3n1 )+(3m2 ,3n2 )+B PB − µδ(m1 +m2 ,n1 +n2 ),(0,0) ) 2 (µ)
=q n1 m2 e22 (m1 + m2 , n1 + n2 ) − q n2 m1 e22 (m1 + m2 , n1 + n2 ). 14
(µ)
(µ)
[e22 (m1 , n1 ), e23 (m2 , n2 )] X X =[− q A2 m1 Q(3m1 ,3n1 )+A PA − q B2 m1 Q(3m1 ,3n1 )+B PB , A∈K1
B∈K−1
− q −m2 n2 µP−(3m2 +1,3n2 +1) − −
X
q n2 B1 +A2 m2 +A2 B1 QA+B+(3m2 +1,3n2 +1) PA PB
A∈K1 B∈K−1
X
′
′
q n2 B1 +B2 m2 +B2 B1 QB ′ +B+(3m2 +1,3n2 +1) PB PB ′ ]
B,B ′ ∈K−1
=
X
q (n2 +A2 +B2 )m1 +n2 B1 +A2 m2 +A2 B1 Q(3m1 ,3n1 )+A+B+(3m2 +1,3n2 +1) PA PB
A∈K1 B∈K−1
−
X
q n2 B1 +(n1 +A2 )m2 +(A2 +n1 )B1 +A2 m1 Q(3m1 ,3n1 )+A+B+(3m2 +1,3n2 +1) PA PB
A∈K1 B∈K−1
− q −m2 n2 +(−n2 −n1 )m1 µP−(3m2 +1,3n2 +1)−(3m1 ,3n1 ) X q n2 (m1 +B1 )+A2 m2 +A2 (m1 +B1 )+B2 m1 Q(3m1 ,3n1 )+A+B+(3m2 +1,3n2 +1) PA PB − A∈K1 B∈K−1
+
X
′
′
′
q (n2 +B2 +B2 )m1 +n2 B1 +B2 m2 +B2 B1 Q(3m1 ,3n1 )+B+B ′ +(3m2 +1,3n2 +1) PB PB ′
B,B ′ ∈K−1
−
X
′
′
′
q n2 (B1 +m1 )+B2 m2 +B2 (B1 +m1 )+B2 m1 Q(3m1 ,3n1 )+B+B ′ +(3m2 +1,3n2 +1) PB PB ′
B,B ′ ∈K−1
−
X
′
′
q n2 B1 +(n1 +B2 )m2 +(B2 +n1 )B1 +B2 m1 Q(3m1 ,3n1 )+B+B ′ +(3m2 +1,3n2 +1) PB PB ′
B,B ′ ∈K−1
(the first term cancels the fourth while the fifth cancels the sixth) =q m2 n1 (−q −(m1 +m2 )(n1 +n2 ) µP−(3m1 +3m2 +1,3n1 +3n2 +1) X q A2 (m1 +m2 )+(n1 +n2 )B1 +A2 B1 QA+B+(3m1 +3m2 +1,3n1 +3n2 +1) PA PB − A∈K1 B∈K−1
−
X
′
′
q B1 (n1 +n2 )+B2 (m1 +m2 )B2 B1 )QB+B ′ +(3m1 +3m2 +1,3n1 +3n2 +1) PB PB ′
B,B ′ ∈K−1 (µ)
=q m2 n1 e23 (m1 + m2 , n1 + n2 ). 15
The following three brackets are easy. (µ)
(µ)
(µ)
(µ)
(µ)
(µ)
[e22 (m1 , n1 ), e31 (m2 , n2 )] = 0 (µ)
[e22 (m1 , n1 ), e32 (m2 , n2 )] = −q n2 m1 e32 (m1 + m2 , n1 + n2 ) [e22 (m1 , n1 ), e33 (m2 , n2 )] = 0.
(µ)
(µ)
[e23 (m1 , n1 ), e23 (m2 , n2 )]
X
=[−q −m1 n1 µP−(3m1 +1,3n1 +1) −
A∈K1 B∈K−1
X
−
′
′
q n1 B1 +B2 m1 +B2 B1 QB+B ′ +(3m1 +1,3n1 +1) PB PB ′ ,
B,B ′ ∈K−1
− q −m2 n2 µP−(3m2 +1,3n2 +1) − −
q n1 B1 +A2 m1 +A2 B1 QA+B+(3m1 +1,3n1 +1) PA PB
X
q n2 B1 +A2 m2 +A2 B1 QA+B+(3m2 +1,3n2 +1) PA PB
A∈K1 B∈K−1
X
′
′
q n2 B1 +B2 m2 +B2 B1 QB+B ′ +(3m2 +1,3n2 +1) PB PB ′ ]
B,B ′ ∈K−1
=
X
µq −m1 n1 +n2 (−m1 −m2 −B1 )+B2 m2 +B2 (−m1 −m2 −B1 ) PB P−(3m1 +1,3n1 +1)−(3m2 +1,3n2 +1)−B
B∈K−1
−
X
µq −m2 n2 +n1 B1 +(−n2 −n1 −B2 )m1 +(−n2 −n1 −B2 )B1 PB P−(3m1 +1,3n1 +1)−(3m2 +1,3n2 +1)−B
B∈K−1
+
X
′
¯
′
¯
′
¯
¯
′
q n1 (B1 +B1 +m2 )+B2 m1 +B2 (B1 +B1 +m2 )+n2 B1 +B2 m2 +B2 B1
¯ ′ ∈K−1 B,B,B
.QB+B ′ +B+(3m P P ′ PB¯ ¯ 1 +1,3n1 +1)+(3m2 +1,3n2 +1) B B X ′ ′ ¯ ¯ ¯ ¯ + q n1 B1 +(B2 +B2 +n2 )m1 +B1 (B2 +B2 +n2 )+n2 B1 +B2 m2 +B2 B1 ¯ ′ ∈K−1 B,B,B
.QB+B ′ +B+(3m P P ′ PB¯ ¯ 1 +1,3n1 +1)+(3m2 +1,3n2 +1) B B X ′ ′ ′ ′ ¯ ¯ ¯ ¯ − q n2 (B1 +B1 +m1 )+B2 m2 +B2 (B1 +B1 +m1 )+n1 B1 +B2 m1 +B2 B1 ¯ ′ ∈K−1 B,B,B
.QB+B ′ +B+(3m P P ′ PB¯ ¯ 1 +1,3n1 +1)+(3m2 +1,3n2 +1) B B 16
X
−
¯
′
¯
′
¯
¯
q n2 B1 +(B2 +B2 +n1 )m2 +(B2 +B2 +n1 )B1 +n1 B1 +B2 m1 +B2 B1
¯ ′ ∈K−1 B,B,B
.QB+B ′ +B+(3m P P ′ PB¯ ¯ 1 +1,3n1 +1)+(3m2 +1,3n2 +1) B B X ′ ′ ′ ′ + q n1 (B1 +B1 +m2 )+A2 m1 +A2 (B1 +B1 +m2 )+n2 B1 +B2 m2 +B2 B1 A∈K1 B,B ′ ∈K−1
.QA+B+B ′ +(3m1 +1,3n1 +1)+(3m2 +1,3n2 +1) PA PB PB ′ X ′ ′ ′ ′ q n1 B1 +(A2 +B2 +n2 )m1 +(A2 +B2 +n2 )B1 +n2 B1 +A2 m2 +A2 B1 + A∈K1 B,B ′ ∈K−1
.QA+B+B ′ +(3m1 +1,3n1 +1)+(3m2 +1,3n2 +1) PA PB PB ′ X ′ ′ ′ ′ − q n2 (B1 +B1 +m1 )+A2 m2 +A2 (B1 +B1 +m1 )+n1 B1 +B2 m1 +B2 B1 A∈K1 B,B ′ ∈K−1
.QA+B+B ′ +(3m1 +1,3n1 +1)+(3m2 +1,3n2 +1) PA PB PB ′ X ′ ′ ′ ′ q n2 B1 +(A2 +B2 +n1 )m2 +(A2 +B2 +n1 )B1 +n1 B1 +A2 m1 +A2 B1 − A∈K1 B,B ′ ∈K−1
.QA+B+B ′ +(3m1 +1,3n1 +1)+(3m2 +1,3n2 +1) PA PB PB ′ = 0
as the first two terms, the third and the sixth, the fourth and the fifth, the seventh and the last term, are negative to each other.
(µ)
(µ)
[e23 (m1 , n1 ), e31 (m2 , n2 )]
X
=[−q −m1 n1 µP−(3m1 +1,3n1 +1) − −
X
q n1 B1 +A2 m1 +A2 B1 QA+B+(3m1 +1,3n1 +1) PA PB
A∈K1 B∈K−1 ′
′
q n1 B1 +B2 m1 +B2 B1 QB+B ′ +(3m1 +1,3n1 +1) PB PB ′ ,
B,B ′ ∈K−1
X
q A1 n2 Q(3m2 −2,3n2 −2)+A PA ]
A∈K1
17
= − q −m1 n1 +(−m1 −m2 )n2 µP−(3m1 +1,3n1 +1)−(3m2 −2,3n2 −2) X ′ ′ ′ − q n1 (m2 +A1 )+A2 m1 +A2 (m2 +A1 )+A1 n2 QA+A′ +(3m2 −2,3n2 −2)+(3m1 +1,3n1 +1) PA PA′ A,A′ ∈K1
+
X
q (A1 +B1 +m1 )n2 +A2 m1 +A2 B1 +n1 B1 QA+B+(3m2 −2,3n2 −2)+(3m1 +1,3n1 +1) PA PB
X
q n1 (m2 +A1 )+B2 m1 +B2 (m2 +A1 )+A1 n2 QA+B+(3m2 −2,3n2 −2)+(3m1 +1,3n1 +1) PA PB
X
q n1 B1 +(A2 +n2 )m1 +(A2 +n2 )B1 +A1 n2 QA+B+(3m2 −2,3n2 −2)+(3m1 +1,3n1 +1) PA PB
A∈K1 B∈K−1
−
A∈K1 B∈K−1
−
A∈K1 B∈K−1
(the third term and the fifth are cancelled) =q n1 m2 (−q −(m1 +m2 )(n1 +n2 ) µP−(3m1 +3m2 −1,3n1 +3n2 −1) X ′ ′ − q (n1 +n2 )A1 +A2 (m1 +m2 )+A2 A1 QA+A′ +(3m2 +3m1 −1,3n2 +3n1 −1) PA PA′ A,A′ ∈K1
−
X
q (n1 +n2 )A1 +B2 (m1 +m2 )+B2 A1 QA+B+(3m2 +3m1 −1,3n2 +3n1 −1) PA PB )
A∈K1 B∈K−1 (µ)
=q n1 m2 e21 (m1 + m2 , n1 + n2 ).
(µ)
(µ)
[e23 (m1 , n1 ), e32 (m2 , n2 )]
X
=[−µq −m1 n1 P−(3m1 +1,3n1 +1) − −
q n1 B1 +A2 m1 +A2 B1 Q(3m1 +1,3n1 +1)+A+B PA PB
A∈K1 B∈K−1
X
′
′
q n1 B1 +B2 m1 +B2 B1 Q(3m1 +1,3n1 +1)+B+B ′ PB PB ′ , Q(3m2 −1,3n2 −1) ]
B,B ′ ∈K−1
= − µq −m1 n1 δ(−m1 ,−n1 ),(m2 ,n2 ) X q n1 m2 +A2 m1 +A2 m2 Q(3m1 +1,3n1 +1)+A+(3m2 −1,3n2 −1) PA − A∈K1
−
X
q n1 m2 +B2 m1 +B2 m2 Q(3m1 +1,3n1 +1)+B+(3m2 −1,3n2 −1) PB
B∈K−1
−
X
′
′
q n1 B1 +n2 m1 +n2 B1 Q(3m1 +1,3n1 +1)+B ′ +(3m2 −1,3n2 −1) PB ′
B ′ ∈K−1
18
X
=q n1 m2 (− −
X
q A2 (m1 +m2 ) Q(3m1 +3m2 ,3n1 +3n2 )+A PA
A∈K1
B∈K−1
1 q B2 (m1 +m2 ) Q(3m1 +3m2 ,3n1 +3n2 )+B PB − µδ(m1 +m2 ,n1 +n2 ),(0,0) ) 2
− q n2 m1 (
X
B∈K−1
1 q B1 (n1 +n2 ) Q(3m1 +3m2 ,3n1 +3n2 )+B PB + µδ(m1 +m2 ,n1 +n2 ),(0,0) ) 2
(µ)
(µ)
=q n1 m2 e22 (m1 + m2 , n1 + n2 ) − q n2 m1 e33 (m1 + m2 , n1 + n2 ). (µ)
(µ)
[e23 (m1 , n1 ), e33 (m2 , n2 )]
X
=[−µq −m1 n1 P−(3m1 +1,3n1 +1) − −
q n1 B1 +A2 m1 +A2 B1 Q(3m1 +1,3n1 +1)+A+B PA PB
A∈K1 B∈K−1
X
′
′
q n1 B1 +B2 m1 +B2 B1 Q(3m1 +1,3n1 +1)+B+B ′ PB PB ′ ,
B,B ′ ∈K−1
X
q B1 n2 Q(3m2 ,3n2 )+B PB ]
B∈K−1
= − q −m1 n1 +(−m1 −m−2)n2 µP−(3m1 +1,3n1 +1)−(3m2 ,3n2 ) X q n1 (m2 +B1 )+A2 m1 +B2 (m2 +B1 )+B1 n2 QA+B+(3m1 +1,3n1 +1)+(3m2 ,3n2 ) PA PB − A∈K1 B∈K−1
X
−
′
′
′
q n1 (m2 +B1 )+B2 m1 +B2 (m2 +B1 )+B1 n2 QB+(3m2 ,3n2 )+B ′ +(3m1 +1,3n1 +1) PB PB ′
B,B ′ ∈K−1
X
−
′
′
q n1 B1 +(n2 +B2 )m1 +(B2 +n2 )B1 +B1 n2 QB+(3m2 ,3n2 )+B ′ +(3m1 +1,3n1 +1) PB PB ′
B,B ′ ∈K−1
X
+
′
′
′
q (B1 +B1 +m1 )n2 +n1 B1 +B2 m1 +B2 B1 QB+(3m2 ,3n2 )+B ′ +(3m1 +1,3n1 +1) PB PB ′
B,B ′ ∈K−1
(the last two terms cancel) =q n1 m2 (−q −(m1 +m2 )(n1 +n2 ) µP−(3m1 +3m2 +1,3n1 +3n2 +1) X q (n1 +n2 )B1 +A2 (m1 +m2 )+A2 B1 QA+B+(3m1 +3m2 +1,3n1 +3n2 +1) PA PB − A∈K1 B∈K−1
−
X
′
′
q (n1 +n2 )B1 +B2 (m1 +m2 )+B2 B1 QB+B ′ +(3m1 +3m2 +1,3n1 +3n2 +1) PB PB ′ )
B,B ′ ∈K−1 (µ)
=q n1 m2 e23 (m1 + m2 , n1 + n2 ). 19
The following five brackets are easy. (µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
(µ)
[e31 (m1 , n1 ), e31 (m2 , n2 )] = 0 [e31 (m1 , n1 ), e32 (m2 , n2 )] = 0 (µ)
[e31 (m1 , n1 ), e33 (m2 , n2 )] = −q m1 n2 e31 (m1 + m2 , n1 + n2 ) [e32 (m1 , n1 ), e32 (m2 , n2 )] = 0 (µ)
[e32 (m1 , n1 ), e33 (m2 , n2 )] = −q m1 n2 e32 (m1 + m2 , n1 + n2 ).
(µ)
(µ)
[e33 (m1 , n1 ), e33 (m2 , n2 )] X X =[ q B1 n1 Q(3m1 ,3n1 )+B PB , q B1 n2 Q(3m2 ,3n2 )+B PB ] B∈K−1
=
X
B∈K−1
q (B1 +m2 )n1 +B1 n2 Q(3m1 ,3n1 )+(3m2 ,3n2 )+B PB
B∈K−1
−
X
q (B1 +m1 )n2 +B1 n1 Q(3m1 ,3n1 )+(3m2 ,3n2 )+B PB
B∈K−1
=q n1 m2
X
q B1 (n1 +n2 Q(3m1 ,3n1 )+(3m2 ,3n2 )+B PB
B∈K−1
− q m1 n2
X
q B1 (n1 +n2 Q(3m1 ,3n1 )+(3m2 ,3n2 )+B PB
B∈K−1 (µ) =q n1 m2 e33 (m1
(µ)
+ m2 , n1 + n2 ) − q n2 m1 e33 (m1 + m2 , n1 + n2 ). (µ)
Next we check the brackets involving D1 (µ)
(µ)
and D2 .
(µ)
[D1 , e11 (m1 , n1 )] X X X 1 =[ A 1 Q A PA + B1 QB PB , q A1 n1 Q(3m1 ,3n1 )+A PA + µδ(m1 ,n1 ),(0,0) ] 2 A∈K1 B∈K−1 A∈K1 X X = (m1 + A1 )q A1 n1 Q(3m1 ,3n1 )+A PA − A1 q A1 n1 Q(3m1 ,3n1 )+A PA A∈K1
=m1 (
X
A∈K1
A∈K1
1 q A1 n1 Q(3m1 ,3n1 )+A PA + µδ(m1 ,n1 ),(0,0) ) 2
(µ)
=m1 e11 (m1 , n1 ). 20
The following two brackets are easy. (µ)
(µ)
(µ)
(µ)
(µ)
(µ)
[D1 , e12 (m1 , n1 )] = m1 e12 (m1 , n1 ) [D1 , e13 (m1 , n1 )] = m1 e13 (m1 , n1 ). (µ)
(µ)
[D1 , e21 (m1 , n1 )] X X =[ A 1 Q A PA + B1 QB PB , −q −m1 n1 µP−(3m1 −1,3n1 −1) A∈K1
−
B∈K−1
X
q
n1 A′1 +A2 m1 +A2 A′1
QA+A′ +(3m1 −1,3n1 −1) PA PA′
′
A,A ∈K1
−
X
q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB ]
A∈K1 B∈K−1
=q −m1 n1 µ(−m1 )P−(3m1 −1,3n1 −1) X ′ ′ (A1 + A′1 + m1 )q n1 A1 +A2 m1 +A2 A1 QA+A′ +(3m1 −1,3n1 −1) PA PA′ − A,A′ ∈K1
+
X
′
′
′
′
A′1 q n1 A1 +A2 m1 +A2 A1 QA+A′ +(3m1 −1,3n1 −1) PA PA′
A,A′ ∈K1
+
X
A1 q n1 A1 +A2 m1 +A2 A1 QA+A′ +(3m1 −1,3n1 −1) PA PA′
A,A′ ∈K1
−
X
(A1 + B1 + m1 )q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB
X
B1 q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB
X
A1 q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB
A∈K1 B∈K−1
+
A∈K1 B∈K−1
+
A∈K1 B∈K−1
=m1 (−q −m1 n1 µP−(3m1 −1,3n1 −1) X ′ ′ − q n1 A1 +A2 m1 +A2 A1 QA+A′ +(3m1 −1,3n1 −1) PA PA′ A,A′ ∈K1
−
X
q n1 A1 +B2 m1 +B2 A1 QA+B+(3m1 −1,3n1 −1) PA PB )
A∈K1 B∈K−1 (µ)
=m1 e21 (m1 , n1 ). 21
(µ)
(µ)
[D1 , e22 (m1 , n1 )] X X X =[ A 1 Q A PA + B1 QB PB , − q A2 m1 Q(3m1 ,3n1 )+A PA A∈K1
B∈K−1
A∈K1
X
1 q B2 m1 Q(3m1 ,3n1 )+B PB − µδ(m1 ,n1 ),(0,0) ] 2 B∈K−1 X X =− (A1 + m1 )q A2 m1 Q(3m1 ,3n1 )+A PA + A1 q A2 m1 Q(3m1 ,3n1 )+A PA −
A∈K1
−
X
A∈K1
(m1 + B1 )q
B 2 m1
Q(3m1 ,3n1 )+B PB +
B∈K−1
=m1 (−
X
B1 q B2 m1 Q(3m1 ,3n1 )+B PB
B∈K−1
X
q A2 m1 Q(3m1 ,3n1 )+A PA −
A∈K1
X
B∈K−1
1 q B2 m1 Q(3m1 ,3n1 )+B PB − µδ(m1 ,n1 ),(0,0) ) 2
(µ) =m1 e22 (m1 , n1 ).
(µ)
(µ)
[D1 , e23 (m1 , n1 )] X X =[ A 1 Q A PA + B1 QB PB , −q −m1 n1 µP−(3m1 +1,3n1 +1) A∈K1
−
X
B∈K−1
q n1 B1 +A2 m1 +A2 B1 QA+B+(3m1 +1,3n1 +1) PA PB
A∈K1 B∈K−1
−
X
′
′
q n1 B1 +B2 m1 +B2 B1 QB+B ′ +(3m1 +1,3n1 +1) PB PB ′ ]
B,B ′ ∈K−1
=q −m1 n1 µ(−m1 )P−(3m1 +1,3n1 +1) X − (A1 + B1 + m1 )q n1 B1 +A2 m1 +A2 B1 QA+B+(3m1 +1,3n1 +1) PA PB A∈K1 B∈K−1
+
X
A1 q n1 B1 +A2 m1 +A2 B1 QA+B+(3m1 +1,3n1 +1) PA PB
X
B1 q n1 B1 +A2 m1 +A2 B1 QA+B+(3m1 +1,3n1 +1) PA PB
A∈K1 B∈K−1
+
A∈K1 B∈K−1
−
X
′
′
(B1 + B1′ + m1 )q n1 B1 +B2 m1 +B2 B1 QB+B ′ +(3m1 +1,3n1 +1) PB PB ′
B,B ′ ∈K−1
22
X
+
′
′
′
′
B1 q n1 B1 +B2 m1 +B2 B1 QB+B ′ +(3m1 +1,3n1 +1) PB PB ′
B,B ′ ∈K−1
X
+
B1′ q n1 B1 +B2 m1 +B2 B1 QB+B ′ +(3m1 +1,3n1 +1) PB PB ′
B,B ′ ∈K−1
=m1 (−q −m1 n1 µP−(3m1 +1,3n1 +1) X − q n1 B1 +A2 m1 +A2 B1 QA+B+(3m1 +1,3n1 +1) PA PB A∈K1 B∈K−1
X
−
′
′
q n1 B1 +B2 m1 +B2 B1 QB+B ′ +(3m1 +1,3n1 +1) PB PB ′ )
B,B ′ ∈K−1 (µ)
=m1 e23 (m1 , n1 ). The following two brackets are easy. (µ)
(µ)
(µ)
(µ)
(µ)
(µ)
[D1 , e31 (m1 , n1 )] = m1 e31 (m1 , n1 ) [D1 , e32 (m1 , n1 )] = m1 e32 (m1 , n1 ). (µ)
(µ)
[D1 , e33 (m1 , n1 )] X X X 1 =[ A 1 Q A PA + B1 QB PB , q B1 n1 Q(3m1 ,3n1 )+B PB + µδ(m1 ,n1 ),(0,0) ] 2 A∈K1 B∈K−1 B∈K−1 X X = (m1 + B1 )q B1 n1 Q(3m1 ,3n1 )+B PB − B1 q B1 n1 Q(3m1 ,3n1 )+B PB B∈K−1
=m1 (
B∈K−1
X
B∈K−1
1 q B1 n1 Q(3m1 ,3n1 )+B PB + µδ(m1 ,n1 ),(0,0) ) 2
(µ)
=m1 e33 (m1 , n1 ) Similarly, we can get (µ)
(µ)
(µ)
[D2 , eij (m1 , n1 )] = n1 eij (m1 , n1 ) for 1 ≤ i, j ≤ 3. Finally, (µ)
(µ)
[D1 , D2 ] X X X X A 1 Q A PA + B1 QB PB , A 2 Q A PA + B2 QB PB ] =[ A∈K1
B∈K−1
A∈K1
B∈K−1
=0 Hence π : gl^ 3 (Cq ) → End(V ) is a Lie algebra homomorphism. 23
§2. Hermitian form for gl^ 3 (Cq )-module. From now on we need to assume that |q| = 1. ^ Define ω : gl^ 3 (Cq ) 7→ gl3 (Cq ) a R-linear map as the following: (2.1) (2.2)
¯ ω(λx) = λω(x), ∀λ ∈ C, x ∈ gl^ 3 (Cq ) ω(Eij (a)) = (−1)i+j Eji (a), a ∈ Cq
(2.3)
ω(ds ) = ds , ω(dt ) = dt , ω(cs ) = cs , ω(ct ) = ct
¯ −n s−m = λq ¯ mn s−m t−n , where R−linear function ¯ : Cq → Cq is defined as λsm tn = λt ¯ is the complex conjugate, for any λ ∈ C, and m, n ∈ Z. and λ Following from [Lemma 3.4, GZ], we have Lemma 2.4. ω is an anti-linear anti-involution of gl^ 3 (Cq ). We simply write π(Eij (r)).v as Eij (r).v, for any v ∈ V, r ∈ Cq . In [GZ], we define a hermitian form on the basis consisting of monomials and then use another basis consisting of iterated module actions on the “highest weight vector ” 1 to determine the condition for the form being positive definite. Here we will use the second basis directly to define the hermitian form which is much simpler. Lemma 2.5. E12 (α1 )E12 (α2 )...E12 (α (β2 )...E32(βl ).1 (We shall call it in S k )E32 (β1 )E32m level (k, l) in W ), here k, l ∈ Z+ {0}, αi = s i tni , i = 1...k, βj = suj tvj , j = 1...l, mi, ni , uj , vj ∈ Z forms a basis for V . Proof. Since fA,B =
Y
A
(m,n) x(3m+1,3n+1) .
(m,n)∈Z2
Y
B
′
′
(m ,n ) x(3m ′ −1,3n′ −1)
(m′ ,n′ )∈Z2
A(m,n) , B(m′ ,n′ ) ∈ Z+ ∪ {0}, where only finitely many A(m,n) , B(m′ ,n′ ) are nonzero, form a basis for V .Q Q B(m′ ,n′ ) A(m,n) Let gA = (m,n)∈Z2 x(3m+1,3n+1) , and hB = (m′ ,n′ )∈Z2 x(3m ′ −1,3n′ −1) . It is similar to [Lemma 4.2, GZ], gA can be written as a linear combination of E12 (α1 )...E12(αk ).1, for P k ≤ (m,n) A(m,n) and hB can be written as a linear combination of E32 (β1 )...E32 (βl ).1, P for l ≤ (m′ ,n′ ) B(m′ ,n′ ) . Since E12 (α)E32 (β).u = E32 (β)E12 (α).u for any u ∈ V , f(A,B) can be written as a linear combination of E12 (α1 )...E12 (αk )E32 (β1 )...E32(βl ).1. Hence E12 (α1 )...E12(αk )E32 (β1 )...E32 (βl ).1 form a basis for V . 24
Let (2.6)
B = {E12 (α1 )...E12 (αk )E32 (β1 )...E32(βl ).1| for all k, l ∈ N, αi , βj ∈ Cq }
be the basis for V . Lemma 2.7. For any v ∈ V , lev(v) = lev(Eii (a).v), i = 1, 2, 3; lev(E12 (a)(v)) = lev(v)+(1, 0); lev(E32 (a).v) = lev(v)+(0, 1); lev(E21 (a).v) = lev(v)−(1, 0) or E21 (a).v = 0 if lev(v)−(1, 0) ∈ / Z2+ ; lev(E23 (a).v) = lev(v)−(0, 1) or E23 (a).v = 0 if lev(v)−(0, 1) ∈ / 2 Z+ , for any 0 6= a ∈ Cq . Proof. We only check those v in the basis B.
E22 (a)E12 (α1 )E12 (α2 )...E12 (αk )E32 (β1 )E32 (β2 )...E32 (βl ).1 =E12 (α1 )E22 (a)E12 (α2 )...E12 (αk )E32 (β1 )E32 (β2 )...E32 (βl ).1 − E12 (α1 a)E12 (α2 )...E12 (αk )E32 (β1 )E32 (β2 )...E32(βl ).1 1 =E12 (α1 )E12 (α2 )...E12 (αk )E32 (β1 )E32 (β2 )...E32 (βl ).( µ)κ(a).1 2 k X E12 (α1 )...E12 (αi a)...E12(αk )E32 (β1 )E32 (β2 )...E32(βl ).1 − i=1
+
l X
E12 (α1 )E12 (α2 )...E12 (αk )E32 (β1 )..E32 (βi a)...E32 (βl ).1,
i=1
so lev(v) = lev(E22 (a).v). It is similar for E11 (a), E33 (a). lev(E12 (a)(v)) = lev(v) + (1, 0) and lev(E32 (a).v) = lev(v) + (0, 1) are the definition of level. For E21 (a).v, we prove by induction on the level of v: E21 (a).v = 0 if lev(v) = (0, n), n ∈ Z+ ∪ {0}: If n = 0, it is obvious that E21 (a).1 = 0. Suppose it is true for n, then E21 (a)E32 (β1 )E32 (β2 )...E32 (βn+1 ).1 =E32 (β1 )E21 (a)E32 (β2 )...E32 (βn+1 ).1 − E31 (β1 a)E32 (β2 )...E32 (βn+1 ).1 = − E32 (β2 )...E32 (βn+1 )E31 (β1 a).1 =0 by induction. 25
Suppose lev(E21 (a).v) = lev(v) − (1, 0) or E21 (a).v = 0 is true for the lev(v) = (m − 1, n), then for v = E12 (b)v ′ with lev(v ′ ) = (m − 1, n), and 0 6= b ∈ Cq E21 (a).E12 (b)v ′ = E12 (b)E21 (a)v ′ + E22 (ab).v ′ − E11 (ba).v ′ . Since lev(E21 (a)v ′ ) = (m − 2, n), lev(E21 (a).E12 (b)v ′ ) = (m − 1, n) or E21 (a).E12(b)v ′ = 0. It is similar for E23 (a). We easily define a contravariant (w.r.t. π, ω) hermitian form on V by defining on the basis B. Assume that µ is a real number, define the conjugate bilinear form on the elements in B by induction on the level: (2.8)
(1, 1) = 1, (1, f ) = 0 if lev(f ) 6= (0, 0)
Suppose for any v ∈ B, (u, v) is defined for any u such that lev(u) = (k ′ , l′ ), with k + l′ = r − 1, if lev(u) = (k, l), with k + l = r, then there exists a u′ such that lev(u′ ) = (k − 1, l), or lev(u′ ) = (k, l − 1), and some a ∈ Cq , such that u = E12 (a)u′ or u = E32 (a)u′ . Define ′
(2.9)
(E12 (a)u′ , v) = (u′ , ω(E12(a))v),
(2.10)
(E32 (a)u′ , v) = (u′ , ω(E32(a))v).
Theorem 2.11. The conjugate bilinear form defined above is a hermitian form on V . Proof. We have to check that (Eij (a)u, v) = (u, ω(Eij (a))v), for 1 ≤ i, j ≤ 3, a ∈ Cq , and (Di .u, v) = (u, ω(Di )v) for i = 1, 2: By the definition, (E12 (a)u, v) = (u, ω(E12(a))v), (E32 (a)u, v) = (u, ω(E32(a))v), and so (E13 (a)u, v) =([E12 (1), E23 (a)]u, v) =(E12 (1)E23 (a)u, v) − (E23 (a)E12 (1)u, v) =(u, ω(E23(a))ω(E12(1))v) − (u, ω(E12(1))ω(E23(a))v) =(u, −ω([E23(a), E12 (1)])v) = (u, ω(E13(a))v). Using induction on the lev(u) to prove (E11 (a).u, v) = (u, ω(E11(a)).v): 26
For any v ∈ B, (E11 (a)1, v) =
1 1 µκ(a)(1, v) = µκ(a)δ1,v . 2 2
Since lev(E11 (a).v) = lev(v) for any v ∈ B, (1, ω(E11(a)).v) = (1, E11 (¯ a).v) =
1 µκ(¯ a)δ1,v . 2
Hence (E11 (a)1, v) = (1, ω(E11(a)).v). Suppose (E11 (a)u, v) = (u, ω(E11 (a)).v) holds true for any lev(u) = (l, k) with l + k = r − 1. For lev(u) = (l, k) with l + k = r, then u = E32 (b).u′ , with lev(u′ ) = (l, k − 1), (E11 (a)E32 (b).u′ , v) =(E32 (b)E11 (a).u′ , v) =(E11 (a).u′ , ω(E32(b)).v) =(u′ , ω(E11 (a))ω(E32(b)).v) =(u′ , ω(E32 (b))ω(E11(a)).v) =(E32 (b)u′ , ω(E11 (a)).v) =(u, ω(E11 (a)).v), or u = E12 (b).u′ with lev(u′ ) = (l − 1, k), (E11 (a)E12 (b).u′ , v) =(E12 (b)E11 (a).u′ , v) + ([E11 (a), E12(b)].u′ , v) =(E11 (a).u′ , ω(E12(b)).v) + (u′ , ω([E11(a), E12 (b)]).v) =(u′ , ω(E11 (a))ω(E12(b)).v) − (u′ , [ω(E11 (a)), ω(E12(b))].v) =(u′ , ω(E12 (b))ω(E11(a)).v) =(E12 (b)u′ , ω(E11 (a)).v) =(u, ω(E11 (a)).v). Thus (E11 (a).u, v) = (u, ω(E11 (a)).v); and (E22 (a).u, v) =([E21 (a), E12(1)].u, v) + (E11 (a)u, v) =(E21 (a)E12 (1).u, v) − (E12 (1)E21 (a).u, v) + (E11 (a)u, v) =(u, ω(E12 (1))ω(E21(a)).v) − (u, ω(E21(a))ω(E12(1)).v) + (u, ω(E11 (a)).v) =(u, ω([E21(a), E12 (1)].v)) + (u, ω(E11 (a)).v) =(u, ω(E22 (a)).v). 27
It is similar for (E33 (a).u, v) = (u, ω(E33 (a)).v). For D1 , D2 , we also prove by induction on the level of u: It is obvious that (D1 .1, v) = 0, for any v ∈ B, so (D1 .1, 1) = (1, D1 .1) = 0, and suppose (1, D1 .v) = 0 is true for those lev(v) = (k ′ , l′ ), with k ′ + l′ = r > 0, then (1, D1 E12 (sm tn ).v) = (1, E12 D1 .v) + (1, m.v) = 0, and (1, D1 E32 (sm tn ).v) = (1, E32 D1 .v) + (1, m.v) = 0. So (D1 .1, v) = (1, D1 .v). Suppose for any v ∈ B, (D1 .u, v) = (u, D1 .v) is true for all lev(u) = (k ′ , l′ ) such that k ′ + l′ = r, then (D1 .E12 (sm tn ).u, v) = (E12 (sm tn )D1 .u, v) + (m.u, v) = (D1 .u, ω(E12(sm tn ))v) + (u, m.v) = (u, D1 ω(E12 (sm tn ))v) + (u, m.v) = (u, ω(E12(sm tn ))D1 .v) = (E12 (sm tn )u, D1 .v). It is similar for (D1 .E32 (sm tn ).u, v) = (E32 (sm tn )u, D1 .v). Hence (D1 .u, v) = (u, D1 .v), and so is (D2 .u, v) = (u, D2 .v). Note that ω(Di ) = Di , i = 1, 2. §3. Conditions for unitarity. In this section we will determine when the hermitian form given last section is positive definite. Let i ∈ N, γ = (γ1 , ..., γs) be the s − partition of i. We denote P ars (i) be the set of all s − partition of i. Let γ ∈ P ars (N ), we say that π1′ × π2′ ∈ SN × SN is equivalent to π1 × π2 ∈ SN × SN , where SN is the permutation group of N letters, if for all z1 , ...zN , w1 , ..., wN ∈ Cq , κ(zπ1′ (1) wπ2′ (1) ...zπ1′ (γ1 ) wπ2′ (γ1 ) )...κ(zπ1′ (γ1 +...γs−1 +1) wπ2′ (γ1 +...+γs−1 +1) ...zπ1′ (N) wπ2′ (N) ) can be obtained from the analogous expression for π1 × π2 only by rotating the variables. (e.g. κ(z1 w1 z2 w2 z3 w3 ) = κ(z3 w3 z1 w1 z2 w2 )). The following lemma is due to [JK2]. Lemma 3.1. Let z1 , z2 , ..zN , w1 , w2 , ..wN ∈ Cq [s±1 , t±1 ] (3.2) 0 0 =
N X
z1 0 0 0 X
z2 0 ... 0 0
zN 0 0 w1
X
0 0 0 w2
0 0 ... 0 wN
0 .1 0
(−1)γ1 −1 (−µ)κ(zπ1 (1) wπ2 (1) ...zπ1 (γ1 ) wπ2 (γ1 ) )
s=1 γ∈P ars (N) [π1 ×π2 ]∈(SN ×SN )(γ)
.(−1)γ2 −1 (−µ)κ(zπ1 (γ1 +1) wπ2 (γ1 +1) ...zπ1 (γ2 ) wπ2 (γ2 ) ). ...(−1)γs −1 (−µ)κ(zπ1 (γ1 +...γs−1 +1) wπ2 (γ1 +...+γs−1 +1) ...zπ1 (N) wπ2 (N) ).1 28
Lemma 3.3. Let ai , ci , bj , dj ∈ Cq , i = 1, ..., m, j = 1, ..., n, and R = (ai cj )m×m , U = R 0 (bi dj )n×n , and set Λ = = (λi,j )(m+n)×(m+n) 0 U (m+n)×(m+n)
(3.4) E21 (a1 )...E21 (am )E23 (b1 )...E23 (bn )E12 (c1 )...E12(cm )E32 (d1 )...E32 (dn ).1 =
m+n X
X
X
(−1)γ1 −1 (−µ)κ(λπ1 (1),π2 (1) ...λπ1 (γ1 )π2 (γ1 ) )
s=1 γ∈P ars (m+n) [π1 ×π2 ]∈(Sm+n ×Sm+n )(γ)
.(−1)γ2 −1 (−µ)κ(λπ1 (γ1 +1),π2 (γ1 +1) ...λπ1 (γ2 ),π2 (γ2 ) ). ...(−1)γs −1 (−µ)κ(λπ1 (γ1 +...γs−1 +1),π2 (γ1 +...+γs−1 +1) ...λπ1 (N),π2 (N) ).1
Remark 3.5. It is easy to see that λi,j in every summand should be from different rows and different columns of Λ. And if the summand of (3.4) contains some λi,j = 0, then this summand is 0. Hence (3.4) in fact is the sum of those λi,j from R and U . Proof. Prove by induction on n: n = 0, (3.4) is just (3.2). Assume (3.4) is true up to n − 1,
E21 (a1 )...E21(am )E23 (b1 )...E23 (bn )E12 (c1 )...E12 (cm )E32 (d1 )...E32 (dn ).1 =E21 (a1 )...E21(am )E23 (b1 )...E23 (bn−1 ) (E12 (c1 )E23 (bn ) − E13 (c1 bn ))E12 (c2 )...E12 (cm )E32 (d1 )...E32 (dn ).1 =E21 (a1 )...E21(am )E23 (b1 )...E23 (bn−1 )E12 (c1 )E23 (bn )E12 (c2 )...E12(cm )E32 (d1 )...E32 (dn ).1 − E21 (a1 )...E21 (am )E23 (b1 )...E23 (bn−1 )E12 (c1 )...E12(cm )(E12 (c1 bn d1 ) + E32 (d1 )E13 (c1 bn ))E32 (d2 )...E32 (dn ).1 =E21 (a1 )...E21(am )E23 (b1 )...E23 (bn−1 )E12 (c1 )E23 (bn )E12 (c2 )...E12(cm )E32 (d1 )...E32 (dn ).1 +
n X
E21 (a1 )...E21 (am )E23 (b1 )...E23 (bn−1 ).
i=1
E12 (−c1 bn di )E12 (c2 )...E12(cm )E32 (d1 )...E\ 32 (di )...E32 (dn ).1 29
=E21 (a1 )...E21(am )E23 (b1 )...E23 (bn−1 )E12 (c1 )E12 (c2 )...E12 (cm )E23 (bn )E32 (d1 )...E32 (dn ).1 +
n X m X
E21 (a1 )...E21 (am )E23 (b1 )...E23 (bn−1 ).
i=1 j=1
E12 (c1 )...E12 (−cj bn di )...E12 (cm )E32 (d1 )...E\ 32 (di )...E32 (dn ).1 =
n X X
E21 (a1 )...E21 (am )E23 (b1 )...E23 (bn−1 )E12 (c1 )E12 (c2 )...E12(cm )
i=1 j>i
E32 (d1 )...E\ 32 (di )...E32 (dj−1 )E32 (−di bn dj − dj bn di )...E32 (dn ).1 +
n X
E21 (a1 )...E21 (am )E23 (b1 )...E23 (bn−1 ).
i=1
E12 (c1 )E12 (c2 )...E12 (cm )E32 (d1 )...E\ 32 (di )...E32 (dn )(−µ)κ(bn di ).1 n m XX + E21 (a1 )...E21 (am )E23 (b1 )...E23 (bn−1 ). i=1 j=1
E12 (c1 )...E12 (−cj bn di )...E12 (cm )E32 (d1 )...E\ 32 (di )...E32 (dn ).1. Using (3.4) is true for n − 1, and expanding it we can get it is also true for n. Lemma 3.6. The hermitian form on different level is 0. Proof. Only need to prove those elements in the basis B. Let u = E12 (a1 )...E12 (am )E32 (b1 )...E32 (bn ).1, v = E12 (c1 )...E12(ck )E32 (d1 )...E32 (dl ).1, and (m, n) 6= (k, l). At first we prove (u, v) = 0 with m = 0: If k = 0, we can suppose n > l, then (u, v) =(E32 (b1 )...E32 (bn ).1, E32 (d1 )...E32 (dl ).1) =((−1)l E23 (d¯l )...E23 (d¯1 )E32 (b1 )...E32 (bn ).1, 1) by Lemma 2.7, lev(E23 (d¯l )...E23 (d¯1 )E32 (b1 )...E32 (bn ).1) = (0, n − l) or E23 (d¯l )...E23(d¯1 )E32 (b1 )...E32(bn ).1 = 0, then (u, v) = 0. For k > 0, (u, v) =(E32 (b1 )...E32 (bn ).1, E12(c1 )...E12 (ck )E32 (d1 )...E32 (dl ).1) =(−E21 (c1 )E32 (b1 )...E32 (bn ).1, E12(c2 )...E12(ck )E32 (d1 )...E32 (dl ).1) 30
then from Lemma 2.7, −E21 (c1 )E32 (b1 )...E32 (bn ).1 = 0, then (u, v) = 0. Without loss of generality, we can assume that m ≤ k, then (u, v) =(E12 (a1 )...E12(am )E32 (b1 )...E32 (bn ).1, E12(c1 )...E12 (ck )E32 (d1 )...E32 (dl ).1) =(E32 (b1 )...E32 (bn ).1, (−1)m E21 (am )...E21 (a1 ).E12 (c1 )...E12 (ck )E32 (d1 )...E32(dl ).1). From Lemma 2.7, lev(E21 (am )...E21(a1 ).E12 (c1 )...E12(ck )E32 (d1 )...E32 (dl ).1) = (k − m, n) or E21 (am )...E21 (a1 ).E12 (c1 )...E12(ck )E32 (d1 )...E32 (dl ).1 = 0, then back to the case m = 0, we get (u, v) = 0. Similarly to [Proposition 4.11, GZ], and together with Lemma 3.3, we have Proposition 3.7. The hermitian form on the same element h in level (m, n) is a polynomial of µ, with the leading term is c(−1)m+n (−µ)m+n = cµm+n with some constant c > 0. Now we can show the following theorem. Theorem 3.8. (π, V ) is unitariazable if and only if µ > 0. Proof. From [Theorem 4.12, GZ], the hermitian form in level (0, n) and (m, 0) is positive definite if and only if needs µ > 0. Define Ta,b (sm1 tn1 sm2 tn2 ...smk tnk ) = sm1 +a tn1 +b sm2 +a tn2 +b ...smk +a tnk +b (a, b ∈ Z). Extend this operator to the linear operator Tg a,b on V by Tg a,b (E12 (α1 )E12 (α2 )...E12 (αk )E32 (β1 )E32 (β2 )...E32 (βl ).1)
=E12 (Ta,b α1 )E12 (Ta,b α2 )...E12 (Ta,b αk )E32 (Ta,b β1 )E32 (Ta,b β2 )...E32(Ta,b βl ).1. Following Lemma 3.3, Tg a,b preserves the hermitian form on V . Denote
Ll,r (M, N ) = Span{E12 (sm1 tn1 )...E12 (sml tnl )E32 (sj1 tk1 )...E32 (sjr tkr ).1 |mi , ni ≧ 0, i = 1..l, jι, kι ≧ 0, l X i=1
mi +
r X
jι ≦ M,
ι=1
r X i=1
31
ni +
r X ι=1
kι ≦ N }.
Since the hermitian form on different level is 0, we will prove the unitarity by induction on the level. For any µ > 0, the form is definite in level (0, n) ([Theorem 4.12,GZ]), and suppose it is definite in level (r, n), for those r < m, and it is not definite in level (m, n). From Proposition 3.7, we know that the hermitian form restrict to this level should be positive definite for µ big enough. Assume it is not positive definite for some µ > 0, then there exist M, N such that the form restrict on Lm,n (M, N ) is not positive definite. From Proposition 3.7, the form on Ll,r (M, N ) varies smoothly with µ. Then we can find a µ0 at which the form is not positive definite, and for all µ > µ0 , it is positive definite. And we write (., .)µ to be the hermitian form at µ. So the radical of the form is non-trivial at µ0 , i.e there exist a nonzero e h ∈ Lm,n (M, N ), such that for any h ∈ Lm,n (M, N ) we have (e h, h)µ0 = 0. Therefore for any arbitrary element hm−1,n in Lm−1,n (M, N ), and any c ∈ C, we have
(E21 (c).e h, hm−1,n )µ0 = 0. Since the form is positive definite in level (m − 1, n), we have E21 (c).e h = 0, for any e e ^ c ∈ C. Replacing h by T−a,−b (h) if necessary, we can write e h=
m X
ai (E12 (1))i xi
i=1
P where xi = E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1 (here it is a finite sum), and αi , βj is the form sl tk and l, k can not both be 0. Let i0 be the smallest one such that ai0 = 6 0, then i0 ≥ 1. 32
Since E21 (c)(E12 (1))i E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1 =(E12 (1))i E21 (c)E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1 + i.(E12 (1))i−1 (E22 (c) − E11 (c))E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1 i.(i − 1) (E12 (1))i−1 E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1 2 i =(E12 (1)) E21 (c)E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1 + (−2c)
+ i.(E12 (1))i−1 ((−2c)(m − i) − n)E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1 + i.(E12 (1))i−1 E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn )(E22 (c) − E11 (c)).1 i.(i − 1) (E12 (1))i−1 E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1 2 =(E12 (1))i E21 (c)E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1 + (−2c)
i.(i − 1) ]. 2 (E12 (1))i−1 E12 (αi+1 )..E12 (αm )E32 (β1 )..E32 (βn ).1, + [ic(−µ0 ) + i((−2c)(m − i) − n) + (−2c)
we have E21 (c)e h = γai0 (E12 (1))i0 −1 xi0 + R
where R contains those with power of E12 (1) greater than i0 − 1, and γ = i0 c(−µ0 ) + i0 ((−2c)(m − i0 ) − n) + (−2c)
i0 .(i0 − 1) = ci0 (−µ0 − (m − i0 ) − (m − 1)). 2
Since m ≥ i0 , i0 ≥ 1, and µ0 ≥ 0, γ 6= 0, contradict with E21 (c)e h = 0. So for any µ > 0, the hermitian form is positive definite. Acknowledgments I am grateful to my supervisors Professors Nantel Bergeron and Yun Gao for their encouragement and support during the preparation of this paper, especially to Professor Yun Gao for drawing my attention to this subject. 33
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Department of Mathematics and Statistics, York University, Toronto, Canada M3J 1P3 E-mail address:
[email protected]
35
